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3 years ago

The first 5 numbers of a sequence are shown below. Which of the following functions produces the sequence with n: (1, 2, ...,

Mathematics

1 answer:

Rasek [7]3 years ago

4 0

Answer:

Step-by-step explanation:

f(n) = 26 + 9(n-1)

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Dan has a favorite number. His number has the digit 8 in the ones place and 5 in tens place. What is Dan's favorite number.

tino4ka555 [31]

Wouldn't it be 58 or 8.5?

4 0

2 years ago

Read 2 more answers

A spherical globe has a diameter of 10 inches. What is the approximate volume of the globe?

adelina 88 [10]

Answer:

523.33 in^3

Step-by-step explanation:

A sphere is a 3-dimensional version of a circle. An example of a sphere is a ball

Volume of a sphere = $\frac{4}{3}$ π $r^{3}$

Where

π = 3.14

r = radius

the diameter is the straight line that passes through the centre of a circle and touches the two edges of the circle.

A radius is half of the diameter

diameter = 2r

Radius = 10 in / 2 = 5 inches

$\frac{4}{3}$ × 3.14 × $5^{3}$ = 523.33 $in^{3}$

6 0

2 years ago

In order to test for the significance of a regression model involving 3 independent variables and 47 observations, the numerator

matrenka [14]

Answer:

The correct option is C

Step-by-step explanation:

From the question we are told that

The number of independent variables is $n = 3$

The number of observation is $z = 47$

Since n is are independent variables then their degree of freedom is 3

The denominator(i.e z) degrees of freedom is evaluated as

$Df(z) = z - (n +1)$

$Df(z) = 47 - (3 +1)$

$Df(z) = 43$

So for the numerator (n) the degree of freedom is Df(n) = 3

So for the denominator(i.e z) the degree of freedom is Df(z) = 43

4 0

3 years ago

Use each model to predict the life expectancy of residents of a country for which the average annual income is $80,000

Mandarinka [93]

The life expectancies of residents of a country for which the average annual income is $80,000 for the three models are 12309.9352, 172.2436 and 4828.1393

The life expectancies of the models are given as:

$y =69.9352 + 0.1530\times (income)$ --- model 1

$y =4.2436 + 0.0021\times (income)$ --- model 2

$y =4.1393 + 0.0603\times (income)$ --- model 3

Given that the average annual income is $80,000;

We simply substitute 80000 for income in the equations of the three models.

So, we have:

Model 1

$y =69.9352 + 0.1530\times (income)$

$y =69.9352 + 0.1530\times 80000$

$y =12309.9352$

Model 2

$y =4.2436 + 0.0021\times (income)$

$y =4.2436 + 0.0021\times 80000$

$y =172.2436$

Model 3

$y =4.1393 + 0.0603\times (income)$

$y =4.1393 + 0.0603\times 80000$

$y =4828.1393$

Hence, the life expectancies are 12309.9352, 172.2436 and 4828.1393